Fibonacci性质

fibonacci.

 
Fibonacci f(n+1)=f(n)+f(n-1) n >= 1
 
1. sum(f(n)^2) = f(n)*f(n+1)
证明如下，面积之和相等

# 绘图代码如下
import matplotlib.pyplot as plt
from matplotlib.path import Path
import matplotlib.patches as patches
import numpy as np

def fib(max):
	a, b = 1, 1;
	while a < max:
		yield a;
		a, b = b, a + b;
		
		
verts = [    
    ]
codes = [
# Path.MOVETO,
# Path.LINETO,
# Path.MOVETO,
# Path.LINETO,
# Path.CLOSEPOLY,
         ]
fig = plt.figure()
ax = fig.add_subplot(211)
'''
 无穷可数数列（有理分数&双射函数f(N*N)->N
'''
# max = 8
# for k in range(1, max + 1):
#     verts.append((0, k))
#     codes.append(Path.MOVETO)
#     verts.append((k, 0))
#     codes.append(Path.LINETO)
#     for i in range(0, 2*k + 2):
#         x = k - i
#         y = i
#         plt.plot([x], [y], 'ro')
#         ax.text(x, y, str(y) + ':' + str(x), fontsize = 10)
# path = Path(verts, codes)
# ax.grid()
# patch = patches.PathPatch(path, facecolor='orange', lw=2)
# ax.add_patch(patch)
# ax.set_xlim(0, max)
# ax.set_ylim(0, max)

'''
 Plot fibonacci block , a prove SUM(f(i)) = f(n)*f(n-1)
'''
ax = fig.add_subplot(111)
max = 30
prev = 1;
sum = 1;
for i, n in enumerate(fib(max)):    
    if (i == 0):
        continue
    if (i % 2):
        rc = plt.Rectangle((prev, 0), n, n)
    else:
        rc = plt.Rectangle((0, prev), n, n)
    prev = n
    sum += n
    rc.set_fill(False) # rectangle , edge size is n = fib(i)
    ax.add_patch(rc)
    ax.set_xlim(0, sum)
    ax.set_ylim(0, sum)
    c_x = rc.get_x() + rc.get_width() / 2  # center .x
    c_y = rc.get_y() + rc.get_height() / 2 # center .y
    ax.text(c_x, c_y, str(n), fontsize = 10)
plt.show()
 
2. f(n)^2+f(n-1)^2 = f(2*n+2)
 归纳法证明如下：
 n=1，2 上述等式成立，假设对于n-1,n上式成立:
 
f(2n) = f(n)^2 + f(n-1)^2          ----(1)
f(2n+2) = f(n + 1)^2 + f(n)^2      ----(2)

下面推导n+1的情况
 
由假设得2-1：
f(2*n+1) = f(n+1)^2-f(n-1)^2     ----(3)
then:
f(2*n+3) = f(2n+1)+f(2n+2)
 = 2f(2n+1) + f(2n)
  = 2[f(n + 1)^2 - f(n-1)^2] + f(n)^2 + f(n-1)^2
   = 2[[f(n + 1) + f(n-1)] * [f(n + 1) - f(n-1)]] + f(n)^2 + f(n-1)^2
    = 2f(n)[f(n + 1) + f(n-1)] + f(n)^2 + f(n-1)^2
     = 2f(n)f(n + 1) + [f(n) + f(n-1)] ^ 2 
      = 2f(n)f(n + 1) + f(n+1)^2 +  f(n) ^2 - f(n) ^2
       = [f(n) + f(n + 1)] ^2 - f(n) ^2
        = f(n+2)^2 - f(n)^2
 
then
 f(2*n+4) =  f(2*n +3) + f(2n+2)
 = f(n+2)^2 - f(n)^2 + f(n + 1)^2 + f(n)^2
 = f(n + 1)^2 + f(n+2)^2
因此对于n + 1，结论成立
 QED.

 
3. f(n)*[f(n-1)+f(n+1)] = f(2*n+1)
 性质2证明已经包含。
 
 
4. 黄金分割性质
 
数列f(n+1)/f(n) 收敛，且趋于一个值1.618([1+sqrt(5)] / 2)
 
假设收敛，let a = f(n), b=f(n+1), b = k*a;
a/(b) = b/(a+b)
 
1/k = k/(1+k)
1+k  =k^2
解k = [1+sqrt(5)] / 2 显然另外一个解[1-sqrt(5)] / 2为负值。